Cabinet drop

Charlie McDonald · Post by **Charlie McDonald** » 19 Jan 2006 11:46 am

Thank you, John.
I must have slipped a digit in my reporting; I have .0017 on string 5 and .0020 on string 6.
A wound string would work beautifully there, as you say. I'm still attached to my G#->F# lower, and went back to a plain string.
From your equation:

... the amount of movement required to raise or lower a string is inversely proportional to the diameter of the core wire.

I see that there is simply not enough travel in the system to accomplish more than a half-step from a small core.

As always, good to have you here, John, and the rest of you fellows--monitoring possible errors from guys who think they're learning faster than they are.
Such as me still haven't fully grasped the slope of the learning curve. It looks like a simple slope at one point, then one realizes that the slope keeps getting steeper.

richard burton · Post by **richard burton** » 19 Jan 2006 1:01 pm

I came up against the problem of having a wound sixth that would raise a half-tone, but wouldn't drop a full-tone, due to lack of changer finger travel on an all-pull steel.

I wouldn't be beat !!

Slightly boring solution for techies only:

The wound sixth would raise more than a full-tone, in other words a half-tone more than required.
By a bit of jiggery-pokery, I held the raise finger, on a spring-loaded rod, at a position approximately a half-tone higher than its normal 'at rest' position.
Thus, I could still raise it a half-tone with the B pedal, and drop it a full-tone with a knee lever.
The knee lever released the spring-loaded rod, allowing the raise finger to go to its 'at rest' position, whilst at the same time pulling the lowering segment of the finger, to achieve a full-tone drop.

Bobby Lee · Post by **Bobby Lee** » 19 Jan 2006 1:58 pm

Just remember that the actual variable on a wound string is the gauge of the core, not the gauge of the complete manufactured string.

Donny Hinson · Post by **Donny Hinson** » 19 Jan 2006 2:23 pm

We've had a lot of general discussions on this problem, and in the end, the general concensus seems to be that a few cents is tolerable. Don't read your digital tuner and sweat trying to correct a problem that's inconsequential. It's only a problem if it affects your playing, and from what I hear listening to most players out there, they have far more to do with "sounding out of tune" than the mechanics of the guitar they're playing! (Myself included.)

Address your big problems first, and save all this chicken-spit stuff to contend with once you've reached full proficiency, and have mastered your part of the playing equation.

my2cents, anyway

David Doggett · Post by **David Doggett** » 19 Jan 2006 9:46 pm

Oh...if you are talking about the 6th string, then forget the 20 wound, and try a 22 wound (as Bobbe suggested) or even 24 wound. The cores are still smaller than a plain 20. There should be less cabinet drop and less B pedal stiffness than a plain 20, and less lower (or raise) travel than a wound 20. I plotted out a graph of standard guages (from typical packages of various makers) versus pitch. There actually were two slightly different curves for wound and unwound strings. A 20 on the plain string curve came out to the same pitch as a 23 or 24 on the wound curve.

The bottom line is that you probably don't want to swap a plain string with exactly the same gauge wound string. The wound one should always be a little bigger, even for the same pitch. You can just keep increasing the gauge of a wound string until you find the gauge that works right for the raises and lowers you want.

A wound string on 6 will be better for intonation, probably because of fewer high overtones and greater flexibility of the smaller core. It will sound more mellow, and maybe better for jazz. But a lot of country pickers prefer the extra twang of a plain 6. I think most mechanical problems with a wound 6 come from using too small a gauge.

If you really want to get squirrely, put the meter on your cabinet rise with the E lower lever, or the D lever of a uni. That's right, you can get cabinet drop with your raises, and cabinet rise with your lowers. When I saw that, I said, oh the hell with it - I'm not good enough to worry about all this s**t. I split the difference of my cabinet drop between the Es with no pedals, and the As with the A and B pedals down, because that is used so frequently. Any other problems are around 5 cents or less, and as mentioned above, that's tolerable - way less than the differences between JI and ET, or between slight changes in bar pressure or placement, etc.

John Macy · Post by **John Macy** » 19 Jan 2006 10:05 pm

Cabinet drop seems to work pretty well on an amp that is intermitent...

JW Day · Post by **JW Day** » 20 Jan 2006 1:44 pm

I suppose my all time favorite singer is Ray Price. I have had the pleasure many times of listening to him live. Never once did I ever hear him say to B.E. or J. Day, that he could hear cabinet drop. Come to think of it, he never mentioned it last summer to Mike Cass when they were up at Nashville,In. And they all played different brands of guitars.

Bill OConnor · Post by **Bill OConnor** » 23 Jan 2006 10:41 pm

CABINET DROP Has anyone thought about adding a third leg at center front? To counter the flex when floor pedals are pressed ?

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Charlie McDonald · Post by **Charlie McDonald** » 24 Jan 2006 3:48 am

I have seen a picture of one of Richard Burton's guitars, and it has a third leg that appears to be a solution for the problem. I'm not sure if that's what it's really for, however.

Duane Reese · Post by **Duane Reese** » 24 Jan 2006 9:44 am

If you are talking about the pedal bar flexing, I don't think that would cause cabinet drop, would it? Unless the pedals at rest are at their upper limit for some reason, the linkage up top sould stay steady and the pedals themselves would move slightly.

Now as far as the cabinet itself being pulled down by the pedal action and bowing slightly... I've never observed such a thing but it seems that it'd take a pretty stiff pedal and a flimsy cabinet to make that a factor.

Any builders looked into this?

richard burton · Post by **richard burton** » 24 Jan 2006 11:17 am

Charlie,
You are correct.

That middle leg adds rigidity, as the stops for the raises are actually under the pedals, so the pedals need to be as solid as possible.

The middle leg also acts as a mounting point for the cable adjusters (just like the brake cable adjusters on a bicycle)

Wayne Franco · Post by **Wayne Franco** » 24 Jan 2006 12:31 pm

Obviously a "male" guitar.

Mark Fasbender · Post by **Mark Fasbender** » 27 Jan 2006 1:38 am

It was stated in an earlier post that a wound 6th would make for a stiffer B pedal. I believe a wound 6th has a longer throw and therefore less tension on the B pedal. It will also move more in tandem with both the 3rd and 5th string assuming the same mechanical advantage(think shobud or P/P) This longer throw would also improve the E drop situation somewhat. Just an observation, but I have read many posts where people say that a thinner string will raise quicker when it is the opposite. It would be difficult to set up a steel to play well and to understand the nature of timing pulls etc. if one were working within the framework of that misconception. To put it simply...........The thinner a string-the further the pull for a given pitch. A heavy string will get there quicker. This applies to plain strings and the core of wound strings.

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Got Twang ?

Mark

Charlie McDonald · Post by **Charlie McDonald** » 27 Jan 2006 3:27 am

From my observations, Mark is right.
In timing pulls, a wound string will raise faster.
A wound sixth didn't increase pedal pressure, but it did resist cabinet flex.

David Doggett · Post by **David Doggett** » 27 Jan 2006 7:13 am

Um...Charlie, your first statement says the opposite of Mark's point. Substituting a wound 0.020 for a plain 0.020, will result in a slower raise with a longer throw, because the core of the wound string is thinner than the plain string. Even an 0.022 or 0.024 wound might have a thinner core, thus slower raise and longer throw, but the difference will be less. I'm not sure at what gauge the wound core will be the same as an 0.020 plain. You have to subtract the gauge of the wrapping from the wound string gauge to figure that out.

Your second statement is true, and is consistent with the thinner core of the wound string and its longer throw.

Ray Minich · Post by **Ray Minich** » 27 Jan 2006 8:21 am

Ed P., For a set of 10 E9th strings tuned to pitch, can you tell us approximately what the total tension is (the sum of the strings, in pounds force) that is trying to curl the body into the shape of a "U"? I'm only asking 'cause I figure you've got an estimate already figured out.
It'll then be interesting to run the beam equations from the free body diagram to figure out what the moment and deflection are theoretically.

Joseph Meditz · Post by **Joseph Meditz** » 27 Jan 2006 9:15 am

Hey Ray,
These are the tensions for D'Addario strings.

1st F#
Plain Steel Locked Twist Ball End
.013
31.2

2nd D#
Plain Steel Locked Twist Ball End
.015
29.4

3rd G
Plain Steel Locked Twist Ball End
.011
28.1

4th E
Plain Steel Locked Twist Ball End
.014
28.7

5th B
Plain Steel Locked Twist Ball End
.017
23.8

6th G#
Plain Steel Locked Twist Ball End
.020
23.3

7th F#
Stainless Steel Round Wound
.026
26.1

8th E
Stainless Steel Round Wound
.030
28.1

9th D
Stainless Steel Round Wound
.034
27.9

10th B
Stainless Steel Round Wound
.038
24.1

Ray Minich · Post by **Ray Minich** » 27 Jan 2006 9:20 am

Thanks Joseph, I'm gonna build a beam model and see what happens. Will let you know. thanks again.
(I gotta get to learnin' ProE Wildfire 2.0 & Ansys FEA, so maybe this'll give me something interesting to model, it'll sure beat boring pipe sections...).[This message was edited by Ray Minich on 27 January 2006 at 09:41 AM.]

Pat Kelly · Post by **Pat Kelly** » 27 Jan 2006 9:55 am

There has been a lot of attention paid to the total amount of tension being applied by the strings. Although this tension is considerable, it would seem to me to be irrelevent to the topic at hand. This tension is static and can in no way contribute to variations in string tuning attributed to applying a raise, in this case, on the sixth string. As a topic for another thread perhaps someone can give consideration to the possibility of installing a tensioning rod under the guitar to counterbalance the tuned strings. However to the matter in hand.

The common thesis seems to be that applying a raise on, say, string 6, we have increased the tension above the guitar thereby tending to flex the guitar body downwards. To my way of thinking this seems to ignore what is going on under the guitar. Surely the increased tension in the string is reciprocated in large part by the increased tension created under the guitar. Depending on mechanical efficiency, a similar extra tension should be evident between the changer and the pedal. The B pedal on my guitar is only eight inches from the sixth string tuning peg.

If my thesis holds any water, the only part of the guitar which should exhibit any symptoms of flexing would be the short section between the pedal and the tuning key (simplified model).

In my opinion, any cabinet drop when it appears would most likely be caused by the downward pressure applied to the pedals, as other contributors have suggested previously.

Maybe Richard Burton is on to something.

Pat

PS: It occurs to me that if someone with cabinet drop took the 6 string raise off the pedal and put it on LKL, that should test this theory out.[This message was edited by Pat Kelly on 27 January 2006 at 10:07 AM.]

Earnest Bovine · Post by **Earnest Bovine** » 27 Jan 2006 10:02 am

It's easy to separate those two causes of "drop".
1. Measure the effect of more tension of the changer by retuning strings by hand.
2. Measure the effect of pulling down on the body by mashing a pdeal that doesn't do anything. For example use a pedal that operates only the other neck, or better yet loosen the tuning nuts for the test.

Ray Minich · Post by **Ray Minich** » 27 Jan 2006 10:45 am

Pat, the loading exerted by the strings acts to pull the guitar body into the shape of a "U". Call it the "preload".
Let's call the body of the guitar a "beam" with a certain cross section and a certain moment of intertia. You can calculate the deflection of the beam due to the "preload" using classic beam deflection formulas.
Knowing this preload and it's deflection at the "tuned" state, we can then calculate the change in total force(s) on the guitar body caused by:
1. The pedal rod downforce at the axle (acting perpendicular to the "beam"
2. The increase or reduction in string force(s) due to raising or lowering.
Then, using the modulus of elasticity for the material you can determine the change in length for the body.
It'll probably be in the millionths of inch per inch lenghth, but it is not zero. Over a 24 inch stringspan the total deflection would add up. It's still not very much but again, it's not zero. That is why it's usually necessary to use an instrument to find the "detuning" effect.
I think Ed Packard may have already performed this exercize.[This message was edited by Ray Minich on 27 January 2006 at 10:47 AM.]

Bobby Lee · Post by **Bobby Lee** » 27 Jan 2006 11:02 am

There has been a lot of attention paid to the total amount of tension being applied by the strings. Although this tension is considerable, it would seem to me to be irrelevent to the topic at hand. This tension is static and can in no way contribute to variations in string tuning attributed to applying a raise, in this case, on the sixth string.

The total amount of tension is not static. It changes every time you push a pedal.

What we call "cabinet drop" is the sum of several forces, including changes in total string tension, downward pressure on the body of the guitar, and deflections that occur on various mechanical parts. The "counter-force" solution is designed to counter-act the cumulative detuning effect of several different problems.

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Ray Minich · Post by **Ray Minich** » 27 Jan 2006 11:36 am

And when you are trying to conceive the force balance system, don't forget that the changer finger mechanism is one of the neatest applications of "toggle" action for mechanical advantage, whereby a slight force applied yields a large force output. Between 179 and 180 degrees you could push two mountains together, or pull them apart.

Pat Kelly · Post by **Pat Kelly** » 27 Jan 2006 12:24 pm

Bob, What I was attempting to say was that the tension in the strings due to standard tuning (upto 400lbs was suggested in one post) is not of concern here. What we are concerned with is the change in tension. The total tension in the strings after a change can be thought of as the initial tension (the tension applied by standard tuning) plus some additional tension applied by the changer to effect the change. Any equation calculating the changes in the guitar as the result of a pedal raise can discount the static component on each side of the equation, that is the tension due to standard tuning. My point was that it doesn't matter whether the string tension in the "rest" state is 40lb or 400lb. That parameter is a "constant" in the calculation.

My main interest in this was that nobody seems to have allowed that the increase in tension in the strings would be significantly compensated for (in terms of bowing the body) by a corresponding increase in tension in the mechanics under the guitar.

PK

Joseph Meditz · Post by **Joseph Meditz** » 27 Jan 2006 1:37 pm

Hi Pat,
I believe I understand your point on what is happening at the changer end of the cabinet.

If you imagine the changer as a vertical see-saw with the strings on top and the pulls underneath, then, to increase the string tension by 10#s requires a tension of 10#s on the pull. At the fullcrum, being a hinge, there is no tendency toward rotation, or as you are saying, the sum of moments is zero. But, at the fullcrum there is a horizontal force of 10 + 10 = 20#a pulling away from the changer. If the fullcrum were located vertically in the cabinet so that it were at the centroidal axis of the inverted U shaped box, then the bending moment on the cabinet would be zero.

If the fulcrum were located away from the centroid, then there will be a resulting bending moment on the cabinet. I guestimate that the cetroid of the U is about 1 inch or so below the top of the mica. But the axis of rotation of the changer rollers is somewhere above the top of the mica, not at the centroidal axis, resulting in a bending moment IMHO. Phew!

Joe